Question Number 170369 by cortano1 last updated on 22/May/22
Answered by som(math1967) last updated on 22/May/22
![sin10=(b/(2a)) 3sin10−4sin^3 10=sin30 ⇒((3b)/(2a)) −((4b^3 )/(8a^3 ))=(1/2) ⇒ ((3a^2 b−b^3 )/(2a^3 ))=(1/2) ⇒3a^2 b−b^3 =a^3 ∴a^3 +b^3 =3a^2 b ar. △ABC=(1/2)×b×(√(a^2 −(b^2 /4))) =(1/2)×b×((√(4a^2 −b^2 ))/2) =(1/4)×(√(4a^2 b^2 −b^4 )) =(1/4)×(√(b(4a^2 b−b^3 ))) =(1/4)×(√(b(4a^2 b−3a^2 b+a^3 ))) [ ∵a^3 +b^3 =3a^2 b] =(1/4)×(√(ba^2 (a+b))) =(1/4)×a×(√(b(a+b))) ((AD)/(sin80))=((BC)/(sin60)) AD=(b/( (√3)))×2×sin80 =(b/( (√3)))×2×((√(4a^2 −b^2 ))/(2a)) =((√(4a^2 b^2 −b^4 ))/( (√3)a))=(√((b(4a^2 b−b^3 ))/3))×(1/a) =(√((b(a^3 +a^2 b))/3))=a×(1/a)(√((b(a+b))/3)) =(√((b(a+b))/3)) cos80=(b/(2a)) tan40=(√((1−cos80)/(1+cos80))) =(√((2a−b)/(2a+b))) =(√((4a^2 −b^2 )/((2a+b)^2 )))=((√(4a^2 −b^2 ))/(2a+b)) ((2r)/b)=tan40 r=(1/2)×((b(√(4a^2 −b^2 )))/(2a+b)) =((√(4a^2 b^2 −b^4 ))/(2(2a+b)))=((√(b(4a^2 b−b^3 )))/(2(2a+b))) =((√(ba^2 (a+b)))/(2(2a+b)))=((a(√(b(a+b))))/(2(2a+b))) (a/(sin80))=2R R=(a/(2sin80))=(a/(2×((√(4a^2 −b^2 ))/(2a)))) =(a^2 /( (√(4a^2 −b^2 ))))=a×(√((a^2 b)/(4a^2 b−b^3 ))) =a(√((a^2 b)/(a^2 b+a^3 )))=a×(√(b/((a+b))))](https://www.tinkutara.com/question/Q170373.png)
$$\:{sin}\mathrm{10}=\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:\mathrm{3}{sin}\mathrm{10}−\mathrm{4}{sin}^{\mathrm{3}} \mathrm{10}={sin}\mathrm{30} \\ $$$$\Rightarrow\frac{\mathrm{3}{b}}{\mathrm{2}{a}}\:−\frac{\mathrm{4}{b}^{\mathrm{3}} }{\mathrm{8}{a}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} }{\mathrm{2}{a}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} ={a}^{\mathrm{3}} \\ $$$$\therefore\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} =\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}} \\ $$$${ar}.\:\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{b}×\sqrt{{a}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{b}×\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{{b}\left(\mathrm{4}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{{b}\left(\mathrm{4}{a}^{\mathrm{2}} {b}−\mathrm{3}{a}^{\mathrm{2}} {b}+{a}^{\mathrm{3}} \right)} \\ $$$$\left[\:\because{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{3}{a}^{\mathrm{2}} {b}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\sqrt{{ba}^{\mathrm{2}} \left({a}+{b}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×{a}×\sqrt{{b}\left({a}+{b}\right)} \\ $$$$\:\frac{{AD}}{{sin}\mathrm{80}}=\frac{{BC}}{{sin}\mathrm{60}} \\ $$$${AD}=\frac{{b}}{\:\sqrt{\mathrm{3}}}×\mathrm{2}×{sin}\mathrm{80} \\ $$$$\:\:=\frac{{b}}{\:\sqrt{\mathrm{3}}}×\mathrm{2}×\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}{a}} \\ $$$$=\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} }}{\:\sqrt{\mathrm{3}}{a}}=\sqrt{\frac{{b}\left(\mathrm{4}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} \right)}{\mathrm{3}}}×\frac{\mathrm{1}}{{a}} \\ $$$$=\sqrt{\frac{{b}\left({a}^{\mathrm{3}} +{a}^{\mathrm{2}} {b}\right)}{\mathrm{3}}}={a}×\frac{\mathrm{1}}{{a}}\sqrt{\frac{{b}\left({a}+{b}\right)}{\mathrm{3}}} \\ $$$$=\sqrt{\frac{{b}\left({a}+{b}\right)}{\mathrm{3}}} \\ $$$${cos}\mathrm{80}=\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:{tan}\mathrm{40}=\sqrt{\frac{\mathrm{1}−{cos}\mathrm{80}}{\mathrm{1}+{cos}\mathrm{80}}} \\ $$$$\:\:=\sqrt{\frac{\mathrm{2}{a}−{b}}{\mathrm{2}{a}+{b}}} \\ $$$$\: \\ $$$$\:=\sqrt{\frac{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\left(\mathrm{2}{a}+{b}\right)^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}{a}+{b}} \\ $$$$\:\frac{\mathrm{2}{r}}{{b}}={tan}\mathrm{40} \\ $$$$\:{r}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{b}\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}{a}+{b}} \\ $$$$=\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{b}^{\mathrm{4}} }}{\mathrm{2}\left(\mathrm{2}{a}+{b}\right)}=\frac{\sqrt{{b}\left(\mathrm{4}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} \right)}}{\mathrm{2}\left(\mathrm{2}{a}+{b}\right)} \\ $$$$=\frac{\sqrt{{ba}^{\mathrm{2}} \left({a}+{b}\right)}}{\mathrm{2}\left(\mathrm{2}{a}+{b}\right)}=\frac{{a}\sqrt{{b}\left({a}+{b}\right)}}{\mathrm{2}\left(\mathrm{2}{a}+{b}\right)} \\ $$$$\:\frac{{a}}{{sin}\mathrm{80}}=\mathrm{2}{R} \\ $$$${R}=\frac{{a}}{\mathrm{2}{sin}\mathrm{80}}=\frac{{a}}{\mathrm{2}×\frac{\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}{\mathrm{2}{a}}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }}={a}×\sqrt{\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}{a}^{\mathrm{2}} {b}−{b}^{\mathrm{3}} }} \\ $$$$={a}\sqrt{\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} {b}+{a}^{\mathrm{3}} }}={a}×\sqrt{\frac{{b}}{\left({a}+{b}\right)}} \\ $$
Commented by Tawa11 last updated on 08/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$